gauravmishra132 wrote:

Hey,

Could you clarify.....if (other,other)is of same kind or different kind?

The answer you have solved above will be true in case of {1,1,1,5,5} but not in case of{1,1,1,5,4}....

here,the number of ways of rolling three 1s in 5 rolls is more than 10....(you could check it manually)

Please clear my doubt...

Hi,

It really doesn't matter what the other two rolls are, as long as they're not 1. The first calculation I made above is for the very specific case where the first three rolls are 1 and the last two rolls are NOT 1. I calculated the probably of this case (that the first three rolls MUST BE 1 and the other two rolls can be

any of 2, 3, 4, 5, or 6 in any combination) to be \(\frac{25}{6^5}\), which is just saying there are \(25\) combinations of {1,1,1,other,other} and \(6^5\) total possibilities (rolling 5 dice, each having 6 sides). Here is where I think you're getting turned around. The 25 combinations are given by:

{1,1,1,2,2} {1,1,1,2,3} {1,1,1,2,4} {1,1,1,2,5} {1,1,1,2,6}

{1,1,1,3,2} {1,1,1,3,3} {1,1,1,3,4} {1,1,1,3,5} {1,1,1,3,6}

{1,1,1,4,2} {1,1,1,4,3} {1,1,1,4,4} {1,1,1,4,5} {1,1,1,4,6}

{1,1,1,5,2} {1,1,1,5,3} {1,1,1,5,4} {1,1,1,5,5} {1,1,1,5,6}

{1,1,1,6,2} {1,1,1,6,3} {1,1,1,6,4} {1,1,1,6,5} {1,1,1,6,6}

But you have to remember that up to this point, we have considered

only one triple-1 case: the case where the first three rolls are all 1. Here is where \(C_5^3\) comes into play. We would like to determine all the different combinations of having three 1s in five rolls, not just the first three rolls. Here, we have \(C_5^3 = 10\) combinations:

{1,1,1,-,-} {1,1,-,1,-} {1,1,-,-,1}

{1,-,1,1,-} {1,-,1,-,1} {1,-,-,1,1}

{-,1,1,1,-} {-,1,1,-,1} {-,1,-,1,1}

{-,-,1,1,1}

Finally, we have to account for the the other numbers that can be the same, not just 1. Hence, we multiply by 6.

I'm far from an expert but this process makes sense to me. Maybe someone else can put it more clearly.